The Brunn-Minkowski inequality can be summarized by stating that the volume, i.e., the Lebesgue measure in \(\mathbb {R}^n\), is (\(1/n\))-concave. More precisely, we have \begin{equation}(1)\quad vol\big((1-\lambda)A+\lambda B\big)^{1/n}\geq(1-\lambda)vol(A)^{1/n}+\lambda vol(B)^{1/n} \end{equation} for all measurable sets \(A, B\) so that \((1-\lambda) A+\lambda B\) is also measurable. It is easy to see that (1) is also true if we exchange \(1/n\) by an arbitrary \(p\leq 1/n\). When working with absolutely continuous measures \(d\mu(x)=f(x)dx\) associated to densities \(f\) with some convexity assumptions, one can also obtain the following Brunn-Minkowski inequality \begin{equation}(2)\quad \mu((1-\lambda) A+\lambda B)^{p}\geq(1-\lambda)\mu(A)^p+\lambda\mu(B)^p \end{equation} for any pair of measurable sets \(A, B\) with \(\mu(A)\mu(B)>0\) and such that \((1-\lambda) A+\lambda B\) is also measurable, where \(p\leq 1/n\) is associated to the ``type of convexity'' of \(f\). The convexity conditions of such density functions \(\)f allow us to understand whether the volume should be the sole measure satisfying the latter inequality or not. Thus, in this talk we will discuss whether, for a given measure on \(\mathbb {R}^n\) (not necessarily absolutely continuous), having an inequality like (2) for a certain (`small') subfamily of sets in \(\mathbb {R}^n\) implies that the measure is (up to a constant) the volume itself. To this respect, we will point out that the constraints \(1/n\geq p>0\), when the support of the measure is the whole \(\mathbb {R}^n\), and \(p=1/n\), when it is an arbitrary open convex set, are both necessary in order to get such a characterization.